Here \(\overset{lim}{x\rightarrow 2}\)\(\frac{3x^{2}-x-10}{x^{2}-4}\) \([\frac{0}{0} form]\)
=\(\overset{lim}{x\rightarrow 2}\)\(\frac{(x-2)(3x+5)}{(x+2)(x-2)}\)
=\(\overset{lim}{x\rightarrow 2}\)\(\frac{(3x+5)}{(x-2)}=\frac{6+5}{2+2}=\frac{11}{4}\).

Here \(\overset{lim}{x\rightarrow -1}\)\(\frac{x^{10}+x^{5}+1}{x-1}\)
putting x = -1
\(=\frac{(-1)^{10}+(-1)^{5}+1}{-1-1}=\frac{1-1+1}{-2}=\frac{0+1}{2}=\frac{-1}{2}\)

Here \(\overset{lim}{x\rightarrow \pi}\) \((x-\frac{22}{7})\) 
putting x = π
= \((\pi-\frac{22}{7})\)

\(Let\quad f(x)={ x }^{ n }+{ ax }^{ n-1 }+{ a }^{ 2 }{ x }^{ n-2 }+...{ a }^{ n-1 }x+{ a }^{ n }\)
\(\therefore f^{\prime}(x)=\frac{d}{d x}\left(x^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n}\right)\)
\(Then,\quad f'(x)=\frac { d }{ dx } (x^{ n })+\frac { d }{ dx } ({ ax }^{ n-1) }+\frac { d }{ dx } ({ a }^{ 2 }{ x }^{ n-2 })+...+\frac { d }{ dx } ({ a }^{ n-1 }x)+\frac { d }{ dx } ({ a }^{ n })\)
\(\text { On using theorem } \frac{d}{d x} x^{n}=n x^{n-1}, \text { we obtain }\)
\(f^{\prime}(x) =n x^{n-1}+a(n-1) x^{n-2}+a^{2}(n-2) x^{n-3}+\ldots+a^{n-1}+a^{n}(0) \)
\(=n x^{n-1}+a(n-1) x^{n-2}+a^{2}(n-2) x^{n-3}+\ldots+a^{n-1}\)

\(f^{\prime}(x)=\frac{(1+\tan x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}} \)
\(f^{\prime}(x)=\frac{(1+\tan x)-x \cdot \frac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}\)
\(\text { Let } g(x)=1+\tan x, \text { Accordingly, } g(x+h)=1+\tan (x+h) \text { . }\)
\(\text { By first principle, }\)
\(g^{\prime}(x) =\lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h} \)
\(=\lim _{h \rightarrow 0}\left[\frac{1+\tan (x+h)-1-\tan x}{h}\right] \)
\(=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}\right] \)
\(=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos (x+h) \cos x}\right]\)
\(=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin (x+h-x)}{\cos (x+h) \cos x}\right] \)
\(=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin h}{\cos (x+h) \cos x}\right] \)
\(=\left(\lim _{h \rightarrow 0} \frac{\sin h}{h}\right) \cdot\left(\lim _{h \rightarrow 0} \frac{1}{\cos (x+h) \cos x}\right) \)
\(=1 \times \frac{1}{\cos ^{2} x}=\sec ^{2} x \)
\(\Rightarrow \frac{d}{d x} (1+\tan x)=\sec ^{2} x\)

\(\text { Let } f(x)=\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}\)
\(\text { By quotient rule }\)
\(f^{\prime}(x) =\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \frac{d}{d x}\left(x^{2}\right)-x^{2} \frac{d}{d x}(\sin x)}{\sin ^{2} x}\right] \)
\(=\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \cdot 2 x-x^{2} \cos x}{\sin ^{2} x}\right] \)
\(=\frac{x \cos \frac{\pi}{4}[2 \sin x-x \cos x]}{\sin ^{2} x}\)

Proof (i) The inequality in (*) says that the function \(\frac{\sin x}{x}\) is sandwiched between the function cos x and the constant function which takes value 1 .
Further, since \(\lim _{x \rightarrow 0} \cos x=1\), we see that the proof of (i) of the theorem is complete by sandwich theorem.
To prove (ii), we recall the trigonometric identity 1 – cos x = 2 sin² \(\left(\frac{x}{2}\right)\)
Then \(\lim _{x \rightarrow 0} \frac{1-\cos x}{x} =\lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\frac{x}{2}\right)}{x}=\lim _{x \rightarrow 0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \sin \left(\frac{x}{2}\right) \\ \)
\( =\lim _{x \rightarrow 0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \lim _{x \rightarrow 0} \sin \left(\frac{x}{2}\right)=1.0=0\)
Observe that we have implicitly used the fact that \(x \rightarrow 0\) is equivalent to \(\frac{x}{2} \rightarrow 0\). This may be justified by putting \(y=\frac{x}{2}\).

\(\text { Let } f(x)=(a x+b)^{n}(c x+d)^{ m}\)
\(\text { By Leibnit product rule, }\)
\(f^{\prime}(x)=(a x+b)^{n} \frac{d}{d x}(c x+d)^{m}+(c x+d)^{m} \frac{d}{d x}(a x+b)^{n}\)
\(\text { Now, let } f_{1}(x)=(c x+d)^{\prime \prime}\)
\(f_{1}(x+h)=(c x+c h+d)^{m}\)
\(f_{1}^{\prime}(x) =\lim _{h \rightarrow 0} \frac{f_{1}(x+h)-f_{1}(x)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{(c x+c h+d)^{m}-(c x+d)^{m}}{h} \)
\(=(c x+d)^{m} \lim _{h \rightarrow 0} \frac{1}{h}\left[\left(1+\frac{c h}{c x+d}\right)^{m}-1\right] \)
\(=(c x+d)^{m} \lim _{h \rightarrow 0} \frac{1}{h}\left[\left(1+\frac{m c h}{(c x+d)}+\frac{m(m-1)}{2} \frac{\left(c^{2} h^{2}\right)}{(c x+d)^{2}}+\ldots\right)-1\right] \)
\(=(c x+d)^{m} \lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{m c h}{(c x+d)}+\frac{m(m-1) c^{2} h^{2}}{2(c x+d)^{2}}+\ldots(\text { Terms containing higher degrees of } h)\right]\)
\(=(c x+d)^{m} \lim _{h \rightarrow 0}\left[\frac{m c}{(c x+d)}+\frac{m(m-1) c^{2} h}{2(c x+d)^{2}}+\ldots\right] \)
\(=(c x+d)^{m}\left[\frac{m c}{c x+d}+0\right] \)
\(=\frac{m c(c x+d)^{m}}{(c x+d)} \)
\(=m c(c x+d)^{m-1} \)
\(\frac{d}{d x}(c x+d)^{m}=m c(c x+d)^{m-1} \)
\(\text { Similarly, } \frac{d}{d x}(a x+b)^{n}=n a(a x+b)^{n-1} \ldots(3)\)
\(\text { Therefore, from (1), (2), and (3), we obtain }\)
\(f^{\prime}(x) =(\alpha x+b)^{n}\left\{m c(c x+d)^{m-1}\right\}+(c x+d)^{m}\left\{n a(a x+b)^{n-1}\right\} \)
\(=(a x+b)^{n-1}(c x+d)^{m-1}[m c(a x+b)+n a(c x+d)]\)

(i) Here f(x) = - x Then if (x + h) = - (x + h) We known that
\(f\left( x \right) =\lim _{ h\rightarrow o }{ \frac { f(x+h)-f(x) }{ h } } \)
\(\Rightarrow f(x)=\lim _{ h\rightarrow o }{ \frac { -(x+h)-(-x) }{ h } } \)
\(=\lim _{ h\rightarrow o }{ \frac { -h }{ h } } =-1\)
Here f(x)=(-x)^-1=-\(-\frac { 1 }{ x } \)
f(x+h)=-\(\frac { 1 }{ x+h } \)
We know that
\(f(x)=\lim _{ h\rightarrow o }{ \frac { f(x+h)-f(x) }{ h } } \)
\(\Rightarrow f\left( x \right) =\lim _{ h\rightarrow o }{ \frac { -\frac { 1 }{ x+h } -\left( -\frac { 1 }{ x } \right) }{ h } } \)
\( =\lim _{ h\rightarrow o }{ \frac { -x+x+h }{ hx(x+h) } } \)
\(=\lim _{ h\rightarrow o }{ \frac { h }{ hx(x+h) } } =\frac { 1 }{ { x }^{ 2 } } \)
(iii) Here f(x)=sin(x+1)
Then f(x+h)=sin(x+h+1)
(iii) We know that
\(f(x)\lim _{ h\rightarrow o }{ \frac { f(x+h)-f(x) }{ h } } \)
\(\Rightarrow f(x)=\lim _{ h\rightarrow o }{ \frac { sin(x+h+1)-sin(x+1) }{ h } } \)
\(=\lim _{ h\rightarrow o }{ \frac { 2cos\left( \frac { 2x+h+2 }{ 2 } \right) sin\frac { h }{ 2 } }{ h } } \)
\(=\lim _{ h\rightarrow o }{ \frac { cos\left( x+1+\frac { h }{ 2 } \right) sin\left( \frac { h }{ 2 } \right) }{ \frac { h }{ 2 } } } \)
\( =cos(x+1)\)
(iv) Here \(f\left( x \right) =cos\left( x-\frac { \pi }{ 8 } \right) \)
Then \(f(x+h)=cos\left( x+h-\frac { \pi }{ 8 } \right) \)
We know that
\(f\left( x \right) =\lim _{ h\rightarrow o }{ \frac { f(x+h)-f(x) }{ h } } \)
\(\Rightarrow f(x)=\lim _{ h\rightarrow o }{ \frac { cos\left( x+h-\frac { \pi }{ 8 } \right) -cos\left( x-\frac { \pi }{ 8 } \right) }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -2sin\left( x-\frac { \pi }{ 8 } +\frac { h }{ 2 } \right) sin\left( \frac { h }{ 2 } \right) }{ \frac { h }{ 2 } } }\)
\(=\lim _{ h\rightarrow o }{ \frac { -sin\left( x-\frac { \pi }{ 8 } +\frac { h }{ 2 } \right) sin\left( \frac { h }{ 2 } \right) }{ \frac { h }{ 2 } } }\)
\(=-sin\left( x-\frac { \pi }{ 8 } \right) \)

(i) Here f(x)=2x-\(\frac{3}{4}\)
∴ \(f^{'}(x)=\frac{d}{dx}(2x-\frac{3}{4})\)
=\(2\frac{d}{dx}(x)-\frac{d}{dx}(\frac{3}{4})\)
=\(2\times 1 -0=2\).
(ii) Here f(x)=(5x³ + 3x - 1) (x - 1)
∴ \(f^{'}(x)=\frac{d}{dx}[(5x^{3}+3x-1)(x-1)]\)
=\((5x^{3}+3x-1)\frac{d}{dx}(x-1)+(x-1)\frac{d}{dx}(5x^{3}+3x-1)\)
=\(5x^{3}+3x-1)\times 1 + (x-1)(15x^{2}+3)\)
=\(5x^{3}+3x-1+15x^{3}+3x-15x^{2}-3\)
=\(20x^{3}-15x^{2}+6x-4\).
(iii) Here f(x)=\(x^{-3} (5+3x)\)
ஃ \(f^{'}(x)=\frac{d}{dx}[x^{-3}(5+3x)]\)
=\(x^{-3}\frac{d}{dx}(5+3x)+(5+3x)\frac{d}{dx}(x^{-3}) \)
=\(x^{-3}\times 3+(5+3x)\times -3x^{-4}\)
=\(\frac{3}{x^{3}}-\frac{3}{x^{4}}(5+3x)\)
=\(\frac{3}{x^{3}}[1-\frac{5+3x}{x}]=\frac{3}{x^{3}}[\frac{x-5-3x}{x}]\)
=\(\frac{-3}{x^4}(5+2x)\).
(iv) Here \(f(x)=x^{5}(3-6x^{-9})\)
∴ \(f^{'}(x)=\frac{d}{dx}[x^{5}(3-6x^{-9})] \)
=\(x^{5}\frac{d}{dx}(3-6x^{-9})+(3-6x^{-9})\frac{d}{dx}(x^{5})\)
=\(x^{5}(54x^{-10})+(3-6x^{-9})\times 5x^{4}\)
=\(54x^{-5}+15x^{4}-30x^{-5}\)
=\(24x^{-5}+15x^{4}\)
=\(\frac{24}{x^{5}}+15x^{4}\).
(v) Here f(x)=x^-4(3-4x^-5)
∴ \(f^{'}(x)=\frac{d}{dx}[x^{-4}(3-4x^{-5})]\)
=\(x^{-4}\frac{d}{dx}(3-4x^{-5})+(3-4x^{-5})\frac{d}{dx}(x^{-4})\)
=\(x^{-4}(20x^{-6})+(3-4x^{-5})(-4x^{-5})\)
=\(20x^{-10}-12x^{-5}+16x^{-10}\)
=\(36x^{-10}-12x^{-5}=\frac{36}{x^{10}}-\frac{12}{x^{5}}\)
(vi) Here, f(x)=\(\frac{2}{x+1}-\frac{x^{2}}{3x-1}\)
ஃ \(f^{'}(x)=\frac{d}{dx}[\frac{2}{x+1}-\frac{x^{2}}{3x-1}]\)
=\(\frac{d}{dx}(\frac{2}{x+1})-\frac{d}{dx}(\frac{x^{2}}{3x-1})\)
=\(\frac{(x+1)\frac{d}{dx}(2)-2\frac{d}{dx}(x+1)}{(x+1)^{2}}\)-\(\frac{(3x-1)\frac{d}{dx}(x^{2}-x^{2}\frac{d}{dx}(3x-1)}{(3x-1)^{2}}\)
=\(\frac{(x+1)\times 0 -2 \times 1}{(x+1)^{2}}-\frac{(3x-1)(2x)-x^{2}\times 3}{(3x-1)^{2}}\)
=\(\frac{-2}{(x+1)^{2}}-\frac{6x^{2}-2x-3x^{2}}{(3x-1)^{2}}\)
=\(\frac{-2}{(x+1)^{2}}-\frac{3x^{2}-2x}{(3x-1)^{2}}\)

We have
\(f^{\prime}(2) =\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0} \frac{3(2+h)-3(2)}{h} \)
\(=\lim _{h \rightarrow 0} \frac{6+3 h-6}{h}=\lim _{h \rightarrow 0} \frac{3 h}{h}=\lim _{h \rightarrow 0} 3=3 .\)
The derivative of the function 3x at x = 2 is 3.

(i) The required limits are all limits of some polynomial functions. Hence the limits are the values of the function at the prescribed points. We have
\(\lim _{x \rightarrow 1}\left[x^{3}-x^{2}+1\right]=1^{3}-1^{2}+1=1 \)
(ii) The required limits are all limits of some polynomial functions. Hence the limits are the values of the function at the prescribed points. We have
\(\lim _{x \rightarrow 3}[x(x+1)]=3(3+1)=3(4)=12\)
(iii) The required limits are all limits of some polynomial functions. Hence the limits are the values of the function at the prescribed points. We have
\(\lim _{x \rightarrow-1}\left[1+x+x^2+\ldots+x^{10}\right]=1+(-1)+(-1)^2+\ldots+(-1)^{10}=1-1+1 \ldots+1=1\)

\(\text { Let } f(x)=\frac{1}{a x^{2}+b x+c}\)
\(\text { By quotient rule, }\)
\(f^{\prime}(x) =\frac{\left(a x^{2}+b x+c\right) \frac{d}{d x}(1)-\frac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}} \)
\(=\frac{\left(a x^{2}+b x+c\right)(0)-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}} \)
\(=\frac{-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}\)

Here f(x)=(ax+b) (cx+d)²
\(\therefore f(x)=\frac { d }{ dx } \left[ (ax+b){ (cx+d) }^{ 2 } \right] \)
\(=(ax+b)\frac { d }{ dx } { (cx+d) }^{ 2 }+{ (cx+d) }^{ 2 }.\frac { d }{ dx } (ax+b)\)
\(=(ax+b)\times 2(cx+d)\times c+2{ (cx+d) }^{ 2 }\times a\)
\(=2c(ax+c)(cx+d){ a(cx+d) }^{ 2 }\)